Đáp án:

\(\begin{array}{l}
Bài\,\,1:\\
1.\\
M = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
2,\\
M = \dfrac{2}{5}\\
Bài\,\,2:\\
1.\\
M = \dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
2.\\
M < 1,\,\,\,\forall \,x > 0\\
Bài\,\,3:\\
1,\\
P = \dfrac{{2\sqrt x  + 1}}{{\sqrt x  + 1}}\\
2,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
{P_{\min }} = 1 \Leftrightarrow x = 0\\
Bài\,\,4:\\
1.\\
P = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
2,\\
\left[ \begin{array}{l}
x = 9\\
x = \dfrac{1}{9}
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
Bài\,\,5:\\
1,\\
M = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
2,\\
M = 0\\
Bài\,\,6:\\
1,\\
M = \dfrac{{x - 13}}{{\sqrt x  + 3}}\\
2,\\
x = 21 + 4\sqrt 5 \\
Bài\,\,7:\\
1,\\
M = \dfrac{{3\sqrt x }}{{\sqrt x  - 3}}\\
2,\\
x \in \left\{ {16;36;144} \right\}\\
Bài\,\,8:\\
1,\\
P = \dfrac{x}{{\sqrt x  - 1}}\\
2,\\
P = 4\\
3,\\
{\sqrt P _{\min }} = 2 \Leftrightarrow x = 4
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Bài\,\,1:\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
1.\\
M = \dfrac{{x + 12}}{{x - 4}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{4}{{\sqrt x  - 2}}\\
 = \dfrac{{x + 12}}{{{{\sqrt x }^2} - {2^2}}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{4}{{\sqrt x  - 2}}\\
 = \dfrac{{x + 12}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{4}{{\sqrt x  - 2}}\\
 = \dfrac{{\left( {x + 12} \right) + \left( {\sqrt x  - 2} \right) - 4.\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x + 12 + \sqrt x  - 2 - 4\sqrt x  - 8}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {x - \sqrt x } \right) + \left( { - 2\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  - 1} \right) - 2.\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
2,\\
x = 9 \Rightarrow M = \dfrac{{\sqrt 9  - 1}}{{\sqrt 9  + 2}} = \dfrac{{3 - 1}}{{3 + 2}} = \dfrac{2}{5}\\
Bài\,\,2:\\
DKXD:\,\,\,x > 0\\
1.\\
M = \dfrac{{x - 2}}{{x + 2\sqrt x }} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x  + 2}}\\
 = \dfrac{{x - 2}}{{\sqrt x \left( {\sqrt x  + 2} \right)}} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x  + 2}}\\
 = \dfrac{{\left( {x - 2} \right) - \left( {\sqrt x  + 2} \right) + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 2 - \sqrt x  - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 4}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} - {2^2}}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
2.\\
M = \dfrac{{\sqrt x  - 2}}{{\sqrt x }} = \dfrac{{\sqrt x }}{{\sqrt x }} - \dfrac{2}{{\sqrt x }} = 1 - \dfrac{2}{{\sqrt x }}\\
x > 0 \Rightarrow \dfrac{2}{{\sqrt x }} > 0\\
 \Rightarrow 1 - \dfrac{2}{{\sqrt x }} < 1,\,\,\,\forall \,x > 0\\
 \Rightarrow M < 1,\,\,\,\forall \,x > 0\\
Bài\,\,3:\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
1,\\
P = \left( {\dfrac{1}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} - 1} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{{{\sqrt x }^2} - {1^2}}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} - 1} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} - 1} \right)\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) + \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt x  + 1 + \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - \sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  - 1} \right)\\
 = \dfrac{{2\sqrt x  + 1}}{{\sqrt x  + 1}}\\
2,\\
P = \dfrac{3}{2}\\
 \Leftrightarrow \dfrac{{2\sqrt x  + 1}}{{\sqrt x  + 1}} = \dfrac{3}{2}\\
 \Leftrightarrow 2.\left( {2\sqrt x  + 1} \right) = 3.\left( {\sqrt x  + 1} \right)\\
 \Leftrightarrow 4\sqrt x  + 2 = 3\sqrt x  + 3\\
 \Leftrightarrow \sqrt x  = 1\\
 \Leftrightarrow x = 1\,\,\,\\
x \ge 0,x \ne 1 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
P = \dfrac{{2\sqrt x  + 1}}{{\sqrt x  + 1}} = \dfrac{{\left( {\sqrt x  + 1} \right) + \sqrt x }}{{\sqrt x  + 1}} = 1 + \dfrac{{\sqrt x }}{{\sqrt x  + 1}}\\
x \ge 0,x \ne 1 \Rightarrow \dfrac{{\sqrt x }}{{\sqrt x  + 1}} \ge 0\\
 \Rightarrow P = 1 + \dfrac{{\sqrt x }}{{\sqrt x  + 1}} \ge 1,\,\,\,\forall x \ge 0,x \ne 1\\
 \Rightarrow {P_{\min }} = 1 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x  + 1}} = 0 \Leftrightarrow \sqrt x  = 0 \Leftrightarrow x = 0\\
Bài\,\,4:\\
DKXD:\,\,\,x > 0\\
1.\\
P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) + \sqrt x .\sqrt x }}{{\sqrt x .\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x }}{{\sqrt x .\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  + 1 + x}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}:\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
2,\\
P = \dfrac{{13}}{3} \Leftrightarrow \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} = \dfrac{{13}}{3}\\
 \Leftrightarrow 3.\left( {x + \sqrt x  + 1} \right) = 13\sqrt x \\
 \Leftrightarrow 3x + 3\sqrt x  + 3 = 13\sqrt x \\
 \Leftrightarrow 3x - 10\sqrt x  + 3 = 0\\
 \Leftrightarrow \left( {3x - 9\sqrt x } \right) + \left( { - \sqrt x  + 3} \right) = 0\\
 \Leftrightarrow 3\sqrt x \left( {\sqrt x  - 3} \right) - \left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left( {\sqrt x  - 3} \right)\left( {3\sqrt x  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - 3 = 0\\
3\sqrt x  - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 3\\
\sqrt x  = \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = \dfrac{1}{9}
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
Bài\,\,5:\\
DKXD:\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 25
\end{array} \right.\\
1,\\
M = \left( {\dfrac{{x + 3\sqrt x }}{{x - 25}} + \dfrac{1}{{\sqrt x  + 5}}} \right):\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \left( {\dfrac{{x + 3\sqrt x }}{{{{\sqrt x }^2} - {5^2}}} + \dfrac{1}{{\sqrt x  + 5}}} \right):\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \left( {\dfrac{{x + 3\sqrt x }}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}} + \dfrac{1}{{\sqrt x  + 5}}} \right):\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \dfrac{{\left( {x + 3\sqrt x } \right) + \left( {\sqrt x  - 5} \right)}}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \dfrac{{x + 4\sqrt x  - 5}}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \dfrac{{\left( {x + 5\sqrt x } \right) + \left( { - \sqrt x  - 5} \right)}}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 5} \right) - \left( {\sqrt x  + 5} \right)}}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 5}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 5} \right)}}{{\left( {\sqrt x  - 5} \right)\left( {\sqrt x  + 5} \right)}}.\dfrac{{\sqrt x  - 5}}{{\sqrt x  + 2}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
2,\\
x = \dfrac{1}{{\sqrt 3  - 1}} - \dfrac{1}{{\sqrt 3  + 1}} = \dfrac{{\left( {\sqrt 3  + 1} \right) - \left( {\sqrt 3  - 1} \right)}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}}\\
 = \dfrac{{\sqrt 3  + 1 - \sqrt 3  + 1}}{{{{\sqrt 3 }^2} - {1^2}}} = \dfrac{2}{{3 - 1}} = \dfrac{2}{2} = 1\\
x = 1 \Rightarrow M = \dfrac{{\sqrt 1  - 1}}{{\sqrt 1  + 2}} = \dfrac{0}{3} = 0\\
Bài\,\,6:\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
1,\\
M = \left( {\dfrac{{x + \sqrt x  - 10}}{{x - 9}} - \dfrac{1}{{\sqrt x  - 3}}} \right):\dfrac{1}{{\sqrt x  - 3}}\\
 = \left( {\dfrac{{x + \sqrt x  - 10}}{{{{\sqrt x }^2} - {3^2}}} - \dfrac{1}{{\sqrt x  - 3}}} \right):\dfrac{1}{{\sqrt x  - 3}}\\
 = \left( {\dfrac{{x + \sqrt x  - 10}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} - \dfrac{1}{{\sqrt x  - 3}}} \right):\dfrac{1}{{\sqrt x  - 3}}\\
 = \dfrac{{\left( {x + \sqrt x  - 10} \right) - \left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{1}{{\sqrt x  - 3}}\\
 = \dfrac{{x + \sqrt x  - 10 - \sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\left( {\sqrt x  - 3} \right)\\
 = \dfrac{{x - 13}}{{\sqrt x  + 3}}\\
2,\\
M = 2\\
 \Leftrightarrow \dfrac{{x - 13}}{{\sqrt x  + 3}} = 2\\
 \Leftrightarrow x - 13 = 2.\left( {\sqrt x  + 3} \right)\\
 \Leftrightarrow x - 13 = 2\sqrt x  + 6\\
 \Leftrightarrow x - 2\sqrt x  - 19 = 0\\
 \Leftrightarrow \left( {x - 2\sqrt x  + 1} \right) - 20 = 0\\
 \Leftrightarrow \left( {{{\sqrt x }^2} - 2\sqrt x .1 + {1^2}} \right) - {\left( {2\sqrt 5 } \right)^2} = 0\\
 \Leftrightarrow {\left( {\sqrt x  - 1} \right)^2} = {\left( {2\sqrt 5 } \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - 1 = 2\sqrt 5 \\
\sqrt x  - 1 =  - 2\sqrt 5 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 2\sqrt 5  + 1\\
\sqrt x  =  - 2\sqrt 5  + 1
\end{array} \right.\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  = 2\sqrt 5  + 1\\
 \Rightarrow x = {\left( {2\sqrt 5  + 1} \right)^2} = {\left( {2\sqrt 5 } \right)^2} + 2.2\sqrt 5 .1 + {1^2} = 20 + 4\sqrt 5  + 1 = 21 + 4\sqrt 5 \\
Bài\,\,7:\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
1,\\
M = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} + \dfrac{{3 - 11\sqrt x }}{{9 - x}}\\
 = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} + \dfrac{{11\sqrt x  - 3}}{{x - 9}}\\
 = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} + \dfrac{{11\sqrt x  - 3}}{{{{\sqrt x }^2} - {3^2}}}\\
 = \dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} + \dfrac{{11\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right) + 11\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  + \sqrt x  + 3 + 11\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x \left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x }}{{\sqrt x  - 3}}\\
2,\\
M = \dfrac{{3\sqrt x }}{{\sqrt x  - 3}} = \dfrac{{\left( {3\sqrt x  - 9} \right) + 9}}{{\sqrt x  - 3}} = \dfrac{{3.\left( {\sqrt x  - 3} \right) + 9}}{{\sqrt x  - 3}} = 3 + \dfrac{9}{{\sqrt x  - 3}}\\
M \in N \Rightarrow \dfrac{9}{{\sqrt x  - 3}} \in Z\\
 \Rightarrow \left( {\sqrt x  - 3} \right) \in Ư\left( 9 \right) = \left\{ { - 9; - 3; - 1;1;3;9} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ { - 9;0;2;4;6;12} \right\}\\
\sqrt x  \ge 0,\sqrt x  \ne 3 \Rightarrow \sqrt x  \in \left\{ {0;2;4;6;12} \right\}\\
 \Rightarrow x \in \left\{ {0;4;16;36;144} \right\}\\
M \in N \Rightarrow M \ge 0 \Rightarrow \dfrac{{3\sqrt x }}{{\sqrt x  - 3}} \ge 0 \Rightarrow \sqrt x  - 3 > 0 \Rightarrow \sqrt x  > 3 \Rightarrow x > 9\\
 \Rightarrow x \in \left\{ {16;36;144} \right\}\\
Bài\,\,8:\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
1,\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\left( {\sqrt x  + 1} \right)}}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{{{\sqrt x }^2} - {1^2}}}} \right):\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\left( {\sqrt x  + 1} \right)}}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}} \right):\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\left( {\sqrt x  + 1} \right)}}} \right)\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  + 1} \right) + \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{2.\left( {\sqrt x  + 1} \right) - \left( {2 - x} \right)}}{{x\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{2\sqrt x  + 2 - 2 + x}}{{x\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{x + 2\sqrt x }}{{x\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{x\left( {\sqrt x  + 1} \right)}}{{x + 2\sqrt x }}\\
 = \dfrac{x}{{\sqrt x  - 1}}\\
2,\\
x = \dfrac{2}{{2 - \sqrt 3 }} - 2\sqrt 3  = \dfrac{{2.\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} - 2\sqrt 3 \\
 = \dfrac{{4 + 2\sqrt 3 }}{{{2^2} - {{\sqrt 3 }^2}}} - 2\sqrt 3  = \dfrac{{4 + 2\sqrt 3 }}{1} - 2\sqrt 3  = 4 + 2\sqrt 3  - 2\sqrt 3  = 4\\
x = 4 \Rightarrow P = \dfrac{4}{{\sqrt 4  - 1}} = \dfrac{4}{{2 - 1}} = \dfrac{4}{1} = 4\\
3,\\
\sqrt P \,\,có\,\,nghĩa \Leftrightarrow P \ge 0 \Leftrightarrow \dfrac{x}{{\sqrt x  - 1}} \ge 0\\
x > 0 \Rightarrow \sqrt x  - 1 > 0 \Rightarrow \sqrt x  > 1 \Rightarrow x > 1\\
P - 4 = \dfrac{x}{{\sqrt x  - 1}} - 4 = \dfrac{{x - 4.\left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{x - 4\sqrt x  + 4}}{{\sqrt x  - 1}} = \dfrac{{x - 2\sqrt x .2 + {2^2}}}{{\sqrt x  - 1}} = \dfrac{{{{\left( {\sqrt x  - 2} \right)}^2}}}{{\sqrt x  - 1}} \ge 0,\,\,\,\forall x > 1\\
 \Rightarrow P \ge 4,\,\,\forall x > 1\\
 \Rightarrow \sqrt P  \ge 2,\,\,\forall x > 1\\
 \Rightarrow {\sqrt P _{\min }} = 2 \Leftrightarrow \dfrac{{{{\left( {\sqrt x  - 2} \right)}^2}}}{{\sqrt x  - 1}} = 0 \Leftrightarrow {\left( {\sqrt x  - 2} \right)^2} = 0 \Leftrightarrow \sqrt x  = 2 \Leftrightarrow x = 4
\end{array}\)